Writing Chemical Formulae — Long Answer Questions (Class 9, CBSE)

Medium Level (Application & Explanation)

Answer:

Valency is the combining power of an element — how many electrons an atom uses to bond with other atoms. It is often the number of electrons gained, lost, or shared to achieve a stable electron configuration.
For main-group elements, valency is related to the number of electrons in the outermost shell.
Example: Sodium (Na) has one valence electron, so valency = 1 → Na⁺.
Oxygen (O) has six outer electrons, needs two more to complete octet, so valency = 2 → O²⁻.
Aluminium (Al) has three outer electrons, so valency = 3 → Al³⁺.
Use valency to balance charges when writing formulae: e.g., Al³⁺ and O²⁻ combine to give Al₂O₃ because 2×(+3) + 3×(−2) = 0.

Answer:

When a compound contains a polyatomic ion (a group of atoms acting as one charged unit), write the ion as a single block. If more than one such block is needed, enclose it in brackets and place the required subscript outside.
Rule: Balance total positive and negative charges.
Example: Calcium hydroxide — Ca²⁺ needs two OH⁻ → Ca(OH)₂ (brackets show two OH groups).
Sodium carbonate — CO₃²⁻ needs two Na⁺ → Na₂CO₃ (carbonate stays as one unit).
Ammonium sulfate — SO₄²⁻ needs two NH₄⁺ → (NH₄)₂SO₄ (brackets show two ammonium ions).
Brackets prevent confusion and show the correct number of atoms in each polyatomic group.

Answer:

Step 1: Write symbols with valencies: Al³⁺ and O²⁻. Mg²⁺ and Cl⁻.
Step 2: Crossover valencies (ignore signs) to become subscripts: For Al and O, cross 3 and 2 → Al₂O₃. For Mg and Cl, cross 2 and 1 → MgCl₂.
Step 3: Check total charge: Al₂O₃ → 2×(+3) + 3×(−2) = 0, balanced. MgCl₂ → +2 + 2×(−1) = 0, balanced.
Step 4: Simplify if possible (remove common factors). Here formulas are already in simplest whole-number ratio.
This method ensures neutral compound by matching positive and negative charges.

Answer:

Read the formula and use subscripts: a subscript after an element or group tells how many atoms of that type are present. If no subscript, it is 1. Multiply subscripts outside brackets by subscripts inside brackets.
For Na₂CO₃: Na = 2, C = 1, O = 3. Total atoms = 2 + 1 + 3 = 6 atoms in one formula unit.
For Al₂(SO₄)₃: Al = 2. Inside (SO₄): S = 1, O = 4. There are 3 such groups, so S = 3×1 = 3, O = 3×4 = 12. Total atoms = 2 (Al) + 3 (S) + 12 (O) = 17 atoms.
Always show work: list each element, apply multipliers, then add.

Answer:

Naming ionic compounds: write the cation (metal) name first, then the anion (non-metal or polyatomic ion). For simple anions, change the ending to “‑ide”; for polyatomic ions use their given names.
CaCl₂: Ca²⁺ = calcium, Cl⁻ = chloride → calcium chloride.
NaNO₃: Na⁺ = sodium, NO₃⁻ = nitrate → sodium nitrate.
Al₂O₃: Al³⁺ = aluminium, O²⁻ = oxide → aluminium oxide.
Metal comes first because ionic compounds form from positively charged cations and negatively charged anions; naming follows this charge order and reflects the compound’s composition clearly.

Answer:

From formulas: XCl₂ means X combines with two Cl atoms, each Cl is Cl⁻ (valency 1). So X must have valency 2 (or charge +2) because it balances two single negative ions. XO shows X combining with one O, and oxygen is O²⁻ (valency 2). XO implies X has a +2 charge to balance O²⁻.
Therefore X is a divalent metal (valency = 2). Common elements with valency 2 include Mg (magnesium), Ca (calcium), Fe²⁺ (iron(II)), Cu²⁺ (copper(II)).
To identify likely element choose typical Group 2 metals (Mg, Ca) as simple candidates because they commonly form stable XCl₂ and XO compounds.

Answer:

Charges: NH₄⁺ (charge +1) and PO₄³⁻ (charge −3). To neutralize charge, you need three NH₄⁺ ions for one PO₄³⁻ ion: 3×(+1) + (−3) = 0.
Write ammonium groups together and use brackets because more than one ammonium unit is needed: (NH₄)₃PO₄.
Name: ammonium phosphate.
Brackets show that each NH₄ group stays intact; subscript 3 applies to the whole ammonium ion, not just N or H separately. This ensures the correct number of atoms and correct charge balance in the compound.

Answer:

Aluminium has valency 3 (Al³⁺) and sulphur has valency 2 (S²⁻). AlS₃ would imply one Al balances three S atoms, giving total charge +3 and 3×(−2)=−6, net −3, not neutral.
Use crossover method: cross valencies 3 (Al) and 2 (S) to make subscripts → Al₂S₃. That means 2×(+3) = +6 and 3×(−2) = −6, total zero.
So Al₂S₃ is the correct empirical formula. The student's AlS₃ fails charge balance; always check that total positive and negative charges sum to zero.

Answer:

Nitrate ion is NO₃⁻, magnesium ion is Mg²⁺. Two NO₃⁻ are needed to balance one Mg²⁺: Mg²⁺ + 2×(NO₃⁻) → Mg(NO₃)₂.
Brackets are necessary because more than one nitrate ion is present; the subscript 2 applies to the whole NO₃ group.
Count atoms: Mg = 1, N = 2 (1×2), O = 6 (3×2). Total atoms = 1 + 2 + 6 = 9 atoms per formula unit.
Brackets avoid writing MgNO₃₂ which would be confusing; Mg(NO₃)₂ clearly shows composition and grouping.

Answer:

Sulfate ion is SO₄²⁻ (charge −2). To neutralize, total positive charge must be +2.
Potassium is K⁺ (charge +1). Two potassium ions are required: K₂SO₄ → 2×(+1) + (−2) = 0. Name: potassium sulphate.
Ammonium is NH₄⁺ (charge +1). Similarly, two ammonium ions are required: (NH₄)₂SO₄ → 2×(+1) + (−2) = 0. Name: ammonium sulphate.
Different cations give different compound names and different molecular compositions, but both need two monovalent cations to balance one divalent sulphate ion. Brackets used for ammonium because the polyatomic ion appears more than once.